Question

A solid sphere of mass 0⋅50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?

Answer 1

Let α be the angular acceleration produced in the sphere.

Rotational equation of motion,

\[F \times R -  f_r  \times R = I\alpha\]

\[\Rightarrow F = \frac{2}{5}mR\alpha + \mu mg........(1)\]

Translational equation of motion,

\[F = ma - \mu mg\]

\[ \Rightarrow a = \frac{\left( F + \mu mg \right)}{m}\]

For pure rolling, we have

\[\alpha = \frac{a}{R}\]

\[\Rightarrow \alpha = \frac{\left( F + \mu mg \right)}{mR}\]

Putting the value of \[\alpha\] in equation (1), we get

\[F = \frac{2}{5}\frac{mR\left( F + \mu mg \right)}{mR} + \mu mg\]

\[ \Rightarrow F = \frac{2}{5}\left( F + \mu mg \right)  \mu mg\]

\[ \Rightarrow F = \frac{2}{5}F + \left( \frac{2}{5} \times \frac{2}{7} \times 0 . 5 \times 10 \right) + \left( \frac{2}{7} \times 0 . 5 \times 10 \right)\]

\[ \Rightarrow \frac{3F}{5} = \frac{4}{7} + \frac{10}{7} = 2\]

\[ \Rightarrow F = \frac{5 \times 2}{3} = \frac{10}{3} = 3 . 3  N\]